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On Liu's book Algebraic Geometry and Arithmetic Curves, he gives a general definition of the projective scheme $\mathbb{P}^n_A$, for $A$ a ring, as $Proj (B)$, for $B=A[x_0,\dots,x_n]$.

Later, he uses a canonical morphism $\pi:\mathbb{P}^n_A\rightarrow \text{Spec }A$. My question is about the nature of this morphism: is $\pi$ induced by the morphism $\text{Spec }(A[x_0,\dots,x_n])\rightarrow \text{Spec }A$, which is induced by the inclusion $A\rightarrow A[x_0,\dots,x_n]$?

Thank you.

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  • $\begingroup$ IIRC, it should be induced by the inclusions of rings $A\rightarrow A[x_1/x_i,\dots,1,\dots,x_n/x_i]$ over the usual affine open cover of $\Bbb P^n=\cup\{x| x_i\ne 0\}$ and then glued together appropriately. $\endgroup$ – PVAL-inactive Nov 24 '14 at 21:40
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    $\begingroup$ @PVAL: I think your comment should be an answer. Also, I'd like to point out that one should think of $\mathbb{P}_A^n$ as a collection of copies of $\mathbb{P}^n$ parameterized by the points of $A$. Then the map $\mathbb{P}_A^n \rightarrow A$ is given by projecting $A \times \mathbb{P}^n \rightarrow A$. [/lowbrow]. $\endgroup$ – RghtHndSd Nov 24 '14 at 21:52
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Recall that $\mathbb{P}^n_A$ is the gluing of affine $n$-spaces over $A$ (namely $\mathrm{Spec} \bigl( A[\frac{x_0}{x_i},\dotsc,\frac{x_n}{x_i}]\bigr) \cong \mathbb{A}^n_A$), and observe that (by the very construction) the gluing isomorphisms at the intersections are morphisms over $A$. It follows that $\mathbb{P}^n_A$ is also an $A$-scheme.

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